Question

If sin 3 $\theta$ = cos ($\theta$-6$°$) then find the value of $\theta$ where (3$\theta$) and ($\theta$-6$°$) are acute angles.

Answer 1

$$\begin{gathered} \mathrm{Given}:\sin 3\theta =\cos (\theta -6°);3\theta \mathrm{and}(\mathrm{\theta }-6°)\mathrm{are}\mathrm{acute}\mathrm{angles} \\ \mathrm{Now},\sin 3\mathrm{\theta }=\cos (\mathrm{\theta }-6°) \\ \Rightarrow \cos (90°-3\mathrm{\theta })=\cos (\mathrm{\theta }-6°)[\because \sin x=\cos (90-x\left)\right] \\ \Rightarrow (90°-3\mathrm{\theta })=(\mathrm{\theta }-6°) \\ \Rightarrow 96°=4\mathrm{\theta } \\ \Rightarrow \mathrm{\theta }=24° \end{gathered}$$